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The value of current i ( t ) shown in the given circuit with the assumption that R = 4 Ω , L = 1 H , C = 0.05 F using Laplace transforms. The value of the current, i ( t ) is i ( t ) = { ( 2 cos 4 t + 9 sin 4 t ) e − 2 t − 2 e − t , 0 < t < 4 ( 2 cos 4 t + 9 sin 4 t ) e − 2 t − ( 2 cos 4 ( t − 4 ) + 9 sin 4 ( t − 4 ) ) e − 2 ( t − 2 ) , t > 4 } _ . Definition used: Laplace transform of derivatives: “The first and second derivatives of f ( t ) satisfy ℒ ( f ′ ) = s ℒ ( f ) − f ( 0 ) ⋯ ( 1 ) ℒ ( f ″ ) = s 2 ℒ ( f ) − s f ( 0 ) − f ′ ( 0 ) ⋯ ( 2 ) Formula (1) holds if f ( t ) is continuous for all t ≥ 0 and satisfies the growth restriction (2) and f ′ ( t ) is piecewise continuous on every finite interval on the semi axis t ≥ 0 . Similarly (2) holds if f ( t ) and f ′ ( t ) are continuous for all t ≥ 0 and satisfy the growth restriction and f ″ ( t ) is piecewise continuous on every finite interval on the semi axis t ≥ 0 .” Unit step function or Heaviside Function: The unit step function or the Heaviside function denoted by u ( t − a ) is defined as follows. u ( t − a ) = { 0 , if t < a 1 , if t > a } Second shifting theorem If f ( t ) has the transform F ( s ) , then the shifted function f ˜ ( t ) = f ( t − a ) u ( t − a ) = { 0 if t < a f ( t − a ) if t > a } has the transform e − a s F ( s ) . That is, if ℒ ( f ( t ) ) = F ( s ) then, ℒ ( f ( t − a ) u ( t − a ) ) = e − a s F ( s ) ⋯ ( 1 ) Calculation: Consider the given conditions as follows. R = 4 Ω , L = 1 H , C = 0.05 F Also, the value of voltage is given as shown below. V = { 34 e − t , 0 < t < 4 0 , t > 4 } It is known that the equation of voltage in an LC circuit is R i ( t ) + L i ′ ( t ) + 1 C ∫ 0 t i ( τ ) d τ = V ( t ) . Substitute the value of voltage in the above equation as shown below. R i ( t ) + L i ′ ( t ) + 1 C ∫ 0 t i ( τ ) d τ = V ( t ) R i ( t ) + L i ′ ( t ) + 1 C ∫ 0 t i ( τ ) d τ = { 34 e − t , 0 < t < 4 0 , t > 4 } Use the definition of the Heaviside step function rewrite the equation of voltage as shown below. v ( t ) = 34 e − t ( 1 − u ( t − 4 ) ) = 34 e − t − 34 e − t + 4 − 4 u ( t − 4 ) = 34 e − t − 34 e − 4 e − ( t − 4 ) u ( t − 4 ) Also, note that the initial current and charge is zero and ℒ ( i ( t ) ) = I ( s ) . Substitute the values R = 4 Ω , L = 1 H , C = 0.05 F and the equation of voltage in the equation R i ( t ) + L i ′ ( t ) + 1 C ∫ 0 t i ( τ ) d τ = V ( t ) L i ″ ( t ) + 1 C i ( t ) = v ′ ( t ) as shown below. 4 i ( t ) + i ′ ( t ) + 1 0.05 ∫ 0 t i ( τ ) d τ = 34 e − t − 34 e − 4 e − ( t − 4 ) u ( t − 4 ) , i ( 0 ) = 0 , i ′ ( 0 ) = 0 Take Laplace transform on both sides of the above equation; apply the second shifting theorem as shown below. ℒ ( i ′ ( t ) + 4 i ( t ) + 20 ∫ 0 t i ( τ ) d τ ) = ℒ ( 34 e − t − 34 e − 4 e − ( t − 4 ) u ( t − 4 ) ) ℒ ( i ′ ( t ) ) + 4 ℒ ( i ( t ) ) + 20 ℒ ( ∫ 0 t i ( τ ) d τ ) = ℒ ( 34 e − t ) − ℒ ( 34 e − 4 e − ( t − 4 ) u ( t − 4 ) ) s I ( s ) + 4 I ( s ) + 20 I ( s ) s = 34 s + 1 − 34 e − 4 s + 1 e − 4 s s 2 + 4 s + 20 s I ( s ) = 34 s + 1 − 34 e − 4 s + 1 e − 4 s I ( s ) = 34 s ( s + 1 ) ( s 2 + 4 s + 20 ) − 34 s e − 4 ( s + 1 ) ( s 2 + 4 s + 20 ) e − 4 s Apply partial fractions in the term s ( s + 1 ) ( s 2 + 4 s + 20 ) as follows. s ( s + 1 ) ( s 2 + 4 s + 20 ) = A s + B s 2 + 4 s + 20 + C s + 1 s ( s + 1 ) ( s 2 + 4 s + 20 ) = ( A s + B ) ( s + 1 ) + C ( s 2 + 4 s + 20 ) ( s + 1 ) ( s 2 + 4 s + 20 ) s = ( A s + B ) ( s + 1 ) + C ( s 2 + 4 s + 20 ) s = A s 2 + A s + B s + B + C s 2 + 4 C s + 20 C Equate the constant terms on both sides as follows. 0 = B + 20 C B = − 20 C Equate the coefficient of s on both sides as shown below. 1 = A + B + 4 C Equate the coefficient of s 2 on both sides as shown below. 0 = A + C A = − C Substitute the value A = − C and B = − 20 C in the equation 1 = A + B + 4 C and solve for C as follows. 1 = A + B + 4 C 1 = − C − 20 C + 4 C 1 = − 17 C − 1 17 = C Therefore, the value of A is 1 17 . Substitute the value of C in B = − 20 C and solve for B as shown below. B = − 20 C = − 20 − 17 = 20 17 Now, substitute the values of A , B and C in the equation s ( s + 1 ) ( s 2 + 4 s + 20 ) = A s + B s 2 + 4 s + 20 + C s + 1 as follows. s ( s + 1 ) ( s 2 + 4 s + 20 ) = A s + B s 2 + 4 s + 20 + C s + 1 = 1 17 ( s 2 + 4 s + 20 ) + 20 17 ( s 2 + 4 s + 20 ) − 1 17 ( s + 1 ) Thus, the value of I ( s ) is obtained as follows. I ( s ) = ( ( 2 s + 40 17 ( s 2 + 4 s + 20 ) − 34 17 ( s + 1 ) ) + e − 4 e − 4 s ( 2 s + 40 17 ( s 2 + 4 s + 20 ) − 34 17 ( s + 1 ) ) ) = ( ( 2 ( s + 20 ) ( s 2 + 4 s + 20 ) − 2 s + 1 ) + ( 2 ( s + 20 ) ( s 2 + 4 s + 20 ) − 2 s + 1 ) e − 4 e − 4 s ) = ( ( ( 2 ( s + 2 ) ( s + 2 ) 2 + 4 2 + 36 ( s + 2 ) 2 + 4 2 ) − 2 s + 1 ) + ( 2 s + 1 − 2 ( s + 2 ) ( s + 2 ) 2 + 4 2 − 36 ( s + 2 ) 2 + 4 2 ) e − 4 e − 4 s ) Now, take inverse Laplace transform on both sides as shown below. ℒ − 1 ( I ( s ) ) = ℒ − 1 ( ( ( 2 ( s + 2 ) ( s + 2 ) 2 + 4 2 + 36 ( s + 2 ) 2 + 4 2 ) − 2 s + 1 ) + ( 2 s + 1 − 2 ( s + 2 ) ( s + 2 ) 2 + 4 2 − 36 ( s + 2 ) 2 + 4 2 ) e − 4 e − 4 s ) = ( ℒ − 1 ( 2 ( s + 2 ) ( s + 2 ) 2 + 4 2 − 36 ( s + 2 ) 2 + 4 2 − 2 s + 1 ) + ℒ − 1 ( ( 2 s + 1 − 2 ( s + 2 ) ( s + 2 ) 2 + 4 2 − 36 ( s + 2 ) 2 + 4 2 ) e − 4 e − 4 s ) ) = ( ℒ − 1 ( 2 ( s + 2 ) ( s + 2 ) 2 + 4 2 ) − ℒ − 1 ( 36 ( s + 2 ) 2 + 4 2 ) − ℒ − 1 ( 2 s + 1 ) + ℒ − 1 ( 2 s + 1 e − 4 e − 4 s ) − ℒ − 1 ( 2 ( s + 2 ) ( s + 2 ) 2 + 4 2 e − 4 e − 4 s ) − ℒ − 1 ( 36 ( s + 2 ) 2 + 4 2 e − 4 e − 4 s ) ) = ( 2 cos 4 t e − 2 t + 9 sin 4 t e − 2 t − 2 e − t + ( 2 e − t − 4 − 2 cos 4 ( t − 4 ) e − 2 ( t − 4 ) − 9 sin 4 ( t − 4 ) e − 2 ( t − 2 ) ) u ( t − 4 ) ) On further simplifications, i ( t ) = ( 2 cos 4 t e − 2 t + 9 sin 4 t e − 2 t − 2 e − t + ( 2 e − t − 4 − 2 cos 4 ( t − 4 ) e − 2 ( t − 4 ) − 9 sin 4 ( t − 4 ) e − 2 ( t − 2 ) ) u ( t − 4 ) ) = ( ( 2 cos 4 t + 9 sin 4 t ) e − 2 t − 2 e − t + ( 2 e − t − ( 2 cos 4 ( t − 4 ) + 9 sin 4 ( t − 4 ) ) e − 2 ( t − 2 ) ) u ( t − 4 ) ) . Obtain the value of i ( t ) in the interval t > 4 as follows. i ( t ) = ( ( 2 cos 4 t + 9 sin 4 t ) e − 2 t − 2 e − t + 2 e − t − ( 2 cos 4 ( t − 4 ) + 9 sin 4 ( t − 4 ) ) e − 2 ( t − 2 ) ) = ( ( 2 cos 4 t + 9 sin 4 t ) e − 2 t − ( 2 cos 4 ( t − 4 ) + 9 sin 4 ( t − 4 ) ) e − 2 ( t − 2 ) ) Thus, the value of i(t) is i ( t ) = { ( 2 cos 4 t + 9 sin 4 t ) e − 2 t − 2 e − t , 0 < t < 4 ( 2 cos 4 t + 9 sin 4 t ) e − 2 t − ( 2 cos 4 ( t − 4 ) + 9 sin 4 ( t − 4 ) ) e − 2 ( t − 2 ) , t > 4 } _ .

The value of current i ( t ) shown in the given circuit with the assumption that R = 4 Ω , L = 1 H , C = 0.05 F using Laplace transforms. The value of the current, i ( t ) is i ( t ) = { ( 2 cos 4 t + 9 sin 4 t ) e − 2 t − 2 e − t , 0 < t < 4 ( 2 cos 4 t + 9 sin 4 t ) e − 2 t − ( 2 cos 4 ( t − 4 ) + 9 sin 4 ( t − 4 ) ) e − 2 ( t − 2 ) , t > 4 } _ . Definition used: Laplace transform of derivatives: “The first and second derivatives of f ( t ) satisfy ℒ ( f ′ ) = s ℒ ( f ) − f ( 0 ) ⋯ ( 1 ) ℒ ( f ″ ) = s 2 ℒ ( f ) − s f ( 0 ) − f ′ ( 0 ) ⋯ ( 2 ) Formula (1) holds if f ( t ) is continuous for all t ≥ 0 and satisfies the growth restriction (2) and f ′ ( t ) is piecewise continuous on every finite interval on the semi axis t ≥ 0 . Similarly (2) holds if f ( t ) and f ′ ( t ) are continuous for all t ≥ 0 and satisfy the growth restriction and f ″ ( t ) is piecewise continuous on every finite interval on the semi axis t ≥ 0 .” Unit step function or Heaviside Function: The unit step function or the Heaviside function denoted by u ( t − a ) is defined as follows. u ( t − a ) = { 0 , if t < a 1 , if t > a } Second shifting theorem If f ( t ) has the transform F ( s ) , then the shifted function f ˜ ( t ) = f ( t − a ) u ( t − a ) = { 0 if t < a f ( t − a ) if t > a } has the transform e − a s F ( s ) . That is, if ℒ ( f ( t ) ) = F ( s ) then, ℒ ( f ( t − a ) u ( t − a ) ) = e − a s F ( s ) ⋯ ( 1 ) Calculation: Consider the given conditions as follows. R = 4 Ω , L = 1 H , C = 0.05 F Also, the value of voltage is given as shown below. V = { 34 e − t , 0 < t < 4 0 , t > 4 } It is known that the equation of voltage in an LC circuit is R i ( t ) + L i ′ ( t ) + 1 C ∫ 0 t i ( τ ) d τ = V ( t ) . Substitute the value of voltage in the above equation as shown below. R i ( t ) + L i ′ ( t ) + 1 C ∫ 0 t i ( τ ) d τ = V ( t ) R i ( t ) + L i ′ ( t ) + 1 C ∫ 0 t i ( τ ) d τ = { 34 e − t , 0 < t < 4 0 , t > 4 } Use the definition of the Heaviside step function rewrite the equation of voltage as shown below. v ( t ) = 34 e − t ( 1 − u ( t − 4 ) ) = 34 e − t − 34 e − t + 4 − 4 u ( t − 4 ) = 34 e − t − 34 e − 4 e − ( t − 4 ) u ( t − 4 ) Also, note that the initial current and charge is zero and ℒ ( i ( t ) ) = I ( s ) . Substitute the values R = 4 Ω , L = 1 H , C = 0.05 F and the equation of voltage in the equation R i ( t ) + L i ′ ( t ) + 1 C ∫ 0 t i ( τ ) d τ = V ( t ) L i ″ ( t ) + 1 C i ( t ) = v ′ ( t ) as shown below. 4 i ( t ) + i ′ ( t ) + 1 0.05 ∫ 0 t i ( τ ) d τ = 34 e − t − 34 e − 4 e − ( t − 4 ) u ( t − 4 ) , i ( 0 ) = 0 , i ′ ( 0 ) = 0 Take Laplace transform on both sides of the above equation; apply the second shifting theorem as shown below. ℒ ( i ′ ( t ) + 4 i ( t ) + 20 ∫ 0 t i ( τ ) d τ ) = ℒ ( 34 e − t − 34 e − 4 e − ( t − 4 ) u ( t − 4 ) ) ℒ ( i ′ ( t ) ) + 4 ℒ ( i ( t ) ) + 20 ℒ ( ∫ 0 t i ( τ ) d τ ) = ℒ ( 34 e − t ) − ℒ ( 34 e − 4 e − ( t − 4 ) u ( t − 4 ) ) s I ( s ) + 4 I ( s ) + 20 I ( s ) s = 34 s + 1 − 34 e − 4 s + 1 e − 4 s s 2 + 4 s + 20 s I ( s ) = 34 s + 1 − 34 e − 4 s + 1 e − 4 s I ( s ) = 34 s ( s + 1 ) ( s 2 + 4 s + 20 ) − 34 s e − 4 ( s + 1 ) ( s 2 + 4 s + 20 ) e − 4 s Apply partial fractions in the term s ( s + 1 ) ( s 2 + 4 s + 20 ) as follows. s ( s + 1 ) ( s 2 + 4 s + 20 ) = A s + B s 2 + 4 s + 20 + C s + 1 s ( s + 1 ) ( s 2 + 4 s + 20 ) = ( A s + B ) ( s + 1 ) + C ( s 2 + 4 s + 20 ) ( s + 1 ) ( s 2 + 4 s + 20 ) s = ( A s + B ) ( s + 1 ) + C ( s 2 + 4 s + 20 ) s = A s 2 + A s + B s + B + C s 2 + 4 C s + 20 C Equate the constant terms on both sides as follows. 0 = B + 20 C B = − 20 C Equate the coefficient of s on both sides as shown below. 1 = A + B + 4 C Equate the coefficient of s 2 on both sides as shown below. 0 = A + C A = − C Substitute the value A = − C and B = − 20 C in the equation 1 = A + B + 4 C and solve for C as follows. 1 = A + B + 4 C 1 = − C − 20 C + 4 C 1 = − 17 C − 1 17 = C Therefore, the value of A is 1 17 . Substitute the value of C in B = − 20 C and solve for B as shown below. B = − 20 C = − 20 − 17 = 20 17 Now, substitute the values of A , B and C in the equation s ( s + 1 ) ( s 2 + 4 s + 20 ) = A s + B s 2 + 4 s + 20 + C s + 1 as follows. s ( s + 1 ) ( s 2 + 4 s + 20 ) = A s + B s 2 + 4 s + 20 + C s + 1 = 1 17 ( s 2 + 4 s + 20 ) + 20 17 ( s 2 + 4 s + 20 ) − 1 17 ( s + 1 ) Thus, the value of I ( s ) is obtained as follows. I ( s ) = ( ( 2 s + 40 17 ( s 2 + 4 s + 20 ) − 34 17 ( s + 1 ) ) + e − 4 e − 4 s ( 2 s + 40 17 ( s 2 + 4 s + 20 ) − 34 17 ( s + 1 ) ) ) = ( ( 2 ( s + 20 ) ( s 2 + 4 s + 20 ) − 2 s + 1 ) + ( 2 ( s + 20 ) ( s 2 + 4 s + 20 ) − 2 s + 1 ) e − 4 e − 4 s ) = ( ( ( 2 ( s + 2 ) ( s + 2 ) 2 + 4 2 + 36 ( s + 2 ) 2 + 4 2 ) − 2 s + 1 ) + ( 2 s + 1 − 2 ( s + 2 ) ( s + 2 ) 2 + 4 2 − 36 ( s + 2 ) 2 + 4 2 ) e − 4 e − 4 s ) Now, take inverse Laplace transform on both sides as shown below. ℒ − 1 ( I ( s ) ) = ℒ − 1 ( ( ( 2 ( s + 2 ) ( s + 2 ) 2 + 4 2 + 36 ( s + 2 ) 2 + 4 2 ) − 2 s + 1 ) + ( 2 s + 1 − 2 ( s + 2 ) ( s + 2 ) 2 + 4 2 − 36 ( s + 2 ) 2 + 4 2 ) e − 4 e − 4 s ) = ( ℒ − 1 ( 2 ( s + 2 ) ( s + 2 ) 2 + 4 2 − 36 ( s + 2 ) 2 + 4 2 − 2 s + 1 ) + ℒ − 1 ( ( 2 s + 1 − 2 ( s + 2 ) ( s + 2 ) 2 + 4 2 − 36 ( s + 2 ) 2 + 4 2 ) e − 4 e − 4 s ) ) = ( ℒ − 1 ( 2 ( s + 2 ) ( s + 2 ) 2 + 4 2 ) − ℒ − 1 ( 36 ( s + 2 ) 2 + 4 2 ) − ℒ − 1 ( 2 s + 1 ) + ℒ − 1 ( 2 s + 1 e − 4 e − 4 s ) − ℒ − 1 ( 2 ( s + 2 ) ( s + 2 ) 2 + 4 2 e − 4 e − 4 s ) − ℒ − 1 ( 36 ( s + 2 ) 2 + 4 2 e − 4 e − 4 s ) ) = ( 2 cos 4 t e − 2 t + 9 sin 4 t e − 2 t − 2 e − t + ( 2 e − t − 4 − 2 cos 4 ( t − 4 ) e − 2 ( t − 4 ) − 9 sin 4 ( t − 4 ) e − 2 ( t − 2 ) ) u ( t − 4 ) ) On further simplifications, i ( t ) = ( 2 cos 4 t e − 2 t + 9 sin 4 t e − 2 t − 2 e − t + ( 2 e − t − 4 − 2 cos 4 ( t − 4 ) e − 2 ( t − 4 ) − 9 sin 4 ( t − 4 ) e − 2 ( t − 2 ) ) u ( t − 4 ) ) = ( ( 2 cos 4 t + 9 sin 4 t ) e − 2 t − 2 e − t + ( 2 e − t − ( 2 cos 4 ( t − 4 ) + 9 sin 4 ( t − 4 ) ) e − 2 ( t − 2 ) ) u ( t − 4 ) ) . Obtain the value of i ( t ) in the interval t > 4 as follows. i ( t ) = ( ( 2 cos 4 t + 9 sin 4 t ) e − 2 t − 2 e − t + 2 e − t − ( 2 cos 4 ( t − 4 ) + 9 sin 4 ( t − 4 ) ) e − 2 ( t − 2 ) ) = ( ( 2 cos 4 t + 9 sin 4 t ) e − 2 t − ( 2 cos 4 ( t − 4 ) + 9 sin 4 ( t − 4 ) ) e − 2 ( t − 2 ) ) Thus, the value of i(t) is i ( t ) = { ( 2 cos 4 t + 9 sin 4 t ) e − 2 t − 2 e − t , 0 < t < 4 ( 2 cos 4 t + 9 sin 4 t ) e − 2 t − ( 2 cos 4 ( t − 4 ) + 9 sin 4 ( t − 4 ) ) e − 2 ( t − 2 ) , t > 4 } _ .

ADVANCED ENGINEERING MATHEMATICS

ADVANCED ENGINEERING MATHEMATICS

10th Edition

ISBN: 2819770198774

Author: Kreyszig

Publisher: WILEY CONS

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Each answer must be justified and all your work should appear. You will be marked on the quality of your explanations. You can discuss the problems with classmates, but you should write your solutions sepa- rately (meaning that you cannot copy the same solution from a joint blackboard, for exam- ple). Your work should be submitted on Moodle, before February 7 at 5 pm. 1. True or false: (a) if E is a subspace of V, then dim(E) + dim(E) = dim(V) (b) Let {i, n} be a basis of the vector space V, where v₁,..., Un are all eigen- vectors for both the matrix A and the matrix B. Then, any eigenvector of A is an eigenvector of B. Justify. 2. Apply Gram-Schmidt orthogonalization to the system of vectors {(1,2,-2), (1, −1, 4), (2, 1, 1)}. 3. Suppose P is the orthogonal projection onto a subspace E, and Q is the orthogonal projection onto the orthogonal complement E. (a) The combinations of projections P+Q and PQ correspond to well-known oper- ators. What are they? Justify your answer. (b) Show…

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ADVANCED ENGINEERING MATHEMATICS

Ch. 6.1 - Prob. 1P Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....

Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Table 6.1. Convert this table to a table for...Ch. 6.1 - Using in Prob. 10, find , where f1(t) = 0 if t ≦...Ch. 6.1 - Table 6.1. Derive formula 6 from formulas 9 and...Ch. 6.1 - Nonexistence. Show that does not satisfy a...Ch. 6.1 - Nonexistence. Give simple examples of functions...Ch. 6.1 - Existence. Show that . [Use (30) in App. 3.1.]...Ch. 6.1 - Change of scale. If and c is any positive...Ch. 6.1 - Inverse transform. Prove that is linear. Hint:...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the shifted data IVPs by the Laplace...Ch. 6.2 - Solve the shifted data IVPs by the Laplace...Ch. 6.2 - Solve the shifted data IVPs by the Laplace...Ch. 6.2 - Solve the shifted data IVPs by the Laplace...Ch. 6.2 - Using (1) or (2), find if f(t) equals: t cos 4t Ch. 6.2 - Using (1) or (2), find if f(t) equals: te−at Ch. 6.2 - Using (1) or (2), find if f(t) equals: cos2 2t Ch. 6.2 - Using (1) or (2), find if f(t) equals: sin2 ωt Ch. 6.2 - Using (1) or (2), find if f(t) equals: sin4 t....Ch. 6.2 - Using (1) or (2), find if f(t) equals: cosh2 t Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.3 - Report on Shifting Theorems. Explain and compare...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Find and sketch or graph f(t) if equals e−3s/(s −...Ch. 6.3 - Prob. 13P Ch. 6.3 - Prob. 14P Ch. 6.3 - Find and sketch or graph f(t) if equals e−3s/s4 Ch. 6.3 - Prob. 16P Ch. 6.3 - Prob. 17P Ch. 6.3 - Using the Laplace transform and showing the...Ch. 6.3 - Using the Laplace transform and showing the...Ch. 6.3 - Prob. 20P Ch. 6.3 - Using the Laplace transform and showing the...Ch. 6.3 - Using the Laplace transform and showing the...Ch. 6.3 - Prob. 23P Ch. 6.3 - Using the Laplace transform and showing the...Ch. 6.3 - Prob. 25P Ch. 6.3 - Prob. 26P Ch. 6.3 - Prob. 27P Ch. 6.3 - Prob. 28P Ch. 6.3 - Prob. 29P Ch. 6.3 - Prob. 30P Ch. 6.3 - Prob. 31P Ch. 6.3 - Prob. 32P Ch. 6.3 - Prob. 33P Ch. 6.3 - Prob. 34P Ch. 6.3 - Prob. 35P Ch. 6.3 - Prob. 36P Ch. 6.3 - Prob. 37P Ch. 6.3 - Prob. 38P Ch. 6.3 - Prob. 39P Ch. 6.3 - Prob. 40P Ch. 6.4 - Prob. 3P Ch. 6.4 - Prob. 4P Ch. 6.4 - Prob. 5P Ch. 6.4 - Prob. 6P Ch. 6.4 - Prob. 7P Ch. 6.4 - Prob. 8P Ch. 6.4 - Prob. 9P Ch. 6.4 - Prob. 11P Ch. 6.4 - Prob. 15P Ch. 6.5 - CONVOLUTIONS BY INTEGRATION Find: Ch. 6.5 - CONVOLUTIONS BY INTEGRATION Find: 2. Ch. 6.5 - CONVOLUTIONS BY INTEGRATION Find: 3. Ch. 6.5 - CONVOLUTIONS BY INTEGRATION Find: 4. Ch. 6.5 - Prob. 5P Ch. 6.5 - Prob. 6P Ch. 6.5 - Prob. 7P Ch. 6.5 - Prob. 8P Ch. 6.5 - Prob. 9P Ch. 6.5 - Prob. 10P Ch. 6.5 - Prob. 11P Ch. 6.5 - Prob. 12P Ch. 6.5 - Prob. 13P Ch. 6.5 - Prob. 14P Ch. 6.5 - CAS EXPERIMENT. Variation of a Parameter. (a)...Ch. 6.5 - Prob. 17P Ch. 6.5 - Prob. 18P Ch. 6.5 - Prob. 19P Ch. 6.5 - Prob. 20P Ch. 6.5 - Prob. 21P Ch. 6.5 - Prob. 22P Ch. 6.5 - Prob. 23P Ch. 6.5 - Prob. 24P Ch. 6.5 - Prob. 25P Ch. 6.5 - Prob. 26P Ch. 6.6 - Prob. 2P Ch. 6.6 - Prob. 3P Ch. 6.6 - Prob. 4P Ch. 6.6 - Prob. 5P Ch. 6.6 - Prob. 6P Ch. 6.6 - Prob. 7P Ch. 6.6 - Prob. 8P Ch. 6.6 - Prob. 9P Ch. 6.6 - Prob. 10P Ch. 6.6 - Prob. 11P Ch. 6.6 - Prob. 14P Ch. 6.6 - Prob. 15P Ch. 6.6 - Prob. 16P Ch. 6.6 - Prob. 17P Ch. 6.6 - Prob. 18P Ch. 6.6 - Prob. 19P Ch. 6.6 - Prob. 20P Ch. 6.7 - Prob. 2P Ch. 6.7 - Prob. 3P Ch. 6.7 - Prob. 4P Ch. 6.7 - Prob. 5P Ch. 6.7 - Prob. 6P Ch. 6.7 - Prob. 7P Ch. 6.7 - Prob. 8P Ch. 6.7 - Prob. 9P Ch. 6.7 - Prob. 10P Ch. 6.7 - Prob. 11P Ch. 6.7 - Prob. 12P Ch. 6.7 - Prob. 13P Ch. 6.7 - Prob. 14P Ch. 6.7 - Prob. 15P Ch. 6.7 - Prob. 16P Ch. 6.7 - Prob. 19P Ch. 6.7 - Prob. 20P Ch. 6 - Prob. 1RQ Ch. 6 - Prob. 2RQ Ch. 6 - Prob. 3RQ Ch. 6 - Prob. 4RQ Ch. 6 - Prob. 5RQ Ch. 6 - When and how do you use the unit step function and...Ch. 6 - If you know f(t) = ℒ−1{F(s)}, how would you find...Ch. 6 - Explain the use of the two shifting theorems from...Ch. 6 - Prob. 9RQ Ch. 6 - Prob. 10RQ Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the inverse transform, indicating the method...Ch. 6 - Prob. 21RQ Ch. 6 - Prob. 22RQ Ch. 6 - Prob. 23RQ Ch. 6 - Prob. 24RQ Ch. 6 - Prob. 25RQ Ch. 6 - Prob. 26RQ Ch. 6 - Prob. 27RQ Ch. 6 - Prob. 28RQ Ch. 6 - Prob. 29RQ Ch. 6 - Prob. 30RQ Ch. 6 - Prob. 31RQ Ch. 6 - Prob. 32RQ Ch. 6 - Prob. 33RQ Ch. 6 - Prob. 34RQ Ch. 6 - Prob. 35RQ Ch. 6 - Prob. 36RQ Ch. 6 - Prob. 37RQ Ch. 6 - Prob. 38RQ Ch. 6 - Prob. 39RQ Ch. 6 - Prob. 40RQ Ch. 6 - Prob. 41RQ Ch. 6 - Prob. 42RQ Ch. 6 - Prob. 43RQ Ch. 6 - Prob. 44RQ Ch. 6 - Prob. 45RQ

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